Dear Eric,
you wrote:
If two or more pairwise defeats have an equal strength and the margins are also equal, they are considered to be equivalent. Starting with the strongest defeat, consider each defeat in sequence with previously kept defeats, if any. If two or more defeats are equivalent, those defeats are considered together with previously kept defeats, if any. If any defeat under consideration is apart of a cycle, it is rejected. If any defeat under consideration is not apart of a cycle, it is kept.
Suppose AB and CD have the same strength. Suppose (a) locking only AB would not create a directed cycle with already locked pairwise defeats, (b) locking only CD would not create a directed cycle with already locked pairwise defeats, but (c) locking both AB and CD simultaneously would create a directed cycle with already locked pairwise defeats.
Then I fear that your formulation could mistakenly be interpreted in such a manner as if both pairwise defeats were rejected.
They should both be rejected as one could not determine which defeat should be kept, unless one would choose randomly between them, but then this would not be a 'deterministic' method.
-- == Eric Gorr ========= http://www.ericgorr.net ========= ICQ:9293199 === "Therefore the considerations of the intelligent always include both benefit and harm." - Sun Tzu == Insults, like violence, are the last refuge of the incompetent... === ---- Election-methods mailing list - see http://electorama.com/em for list info
